The analysis of the molecules \(\text{NH}_3\) and \(\text{NF}_3\) is as follows:
Step 1: Structure and dipole moment of \(\text{NH}_3\)
\(\text{NH}_3\) has a pyramidal shape due to the presence of one lone pair on the nitrogen atom.
- The dipole moments of the \(\text{N–H}\) bonds and the lone pair point in the same direction, leading to a higher resultant dipole moment.
Step 2: Structure and dipole moment of \(\text{NF}_3\)
\(\text{NF}_3\) also has a pyramidal shape, but the \(\text{N–F}\) bonds are highly electronegative.
- The dipole moment of the lone pair on nitrogen is opposite to the resultant dipole moment of the \(\text{N–F}\) bonds, which reduces the overall dipole moment.
Step 3: Comparison of dipole moments
- The dipole moment of \(\text{NH}_3\) is approximately \(1.47 \, \text{D}\), while that of \(\text{NF}_3\) is approximately \(0.80 \, \text{D}\).
- This confirms that \(\text{NH}_3\) has a greater dipole moment than \(\text{NF}_3\).
Step 4: Validating the statements
- Assertion (A): True, because \(\text{NH}_3\) has a higher dipole moment than \(\text{NF}_3\).
- Reason (R): True, as the lone pair’s dipole in \(\text{NH}_3\) aligns with the bond dipoles, while in \(\text{NF}_3\), it opposes them.
- \((R)\) is the correct explanation of \((A)\).
Final Answer: (1).
List-I (Compound / Species) | List-II (Shape / Geometry) |
---|---|
(A) \(SF_4\) | (I) Tetrahedral |
(B) \(BrF_3\) | (II) Pyramidal |
(C) \(BrO_{3}^{-}\) | (III) See saw |
(D) \(NH^{+}_{4}\) | (IV) Bent T-shape |
List - I | List - II |
---|---|
(A) ICl | (IV) Linear |
(B) ICl3 | (I) T-Shape |
(C) ClF5 | (II) Square pyramidal |
(D) IF7 | (III) Pentagonal bipyramidal |
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: